Problem Description

You are given a string s and an integer k. Your task is to count the total number of substrings where at least one character appears at least k times.

A substring is a contiguous sequence of characters within the string. For each valid substring, there must be at least one character that occurs k or more times within that substring.

For example, if s = "aaab" and k = 2, the substring "aaa" would be valid because the character 'a' appears 3 times (which is ≥ 2). Similarly, "aaab" would also be valid because 'a' appears 3 times.

The solution uses a sliding window approach with two pointers. It enumerates each position as a potential right endpoint of substrings. For each right endpoint position, it maintains a left pointer that moves rightward whenever the current window contains a character appearing at least k times. The key insight is that once a window [l, r] contains a character appearing at least k times, all substrings ending at r with starting positions from 0 to l-1 are also valid (since they would contain the entire window [l, r] as a subset). This allows us to efficiently count valid substrings by adding l to our answer for each position of r.

Intuition

The key observation is that if a substring contains a character that appears at least k times, then any longer substring that contains this substring will also satisfy the condition. This monotonic property suggests we can use a sliding window approach.

Let's think about this differently: instead of directly counting valid substrings, we can count them by fixing the right endpoint and determining how many valid left endpoints exist.

For any position r as the right endpoint, we want to find the rightmost position l such that the substring s[l...r] contains at least one character appearing k times. Once we find this position l, all substrings ending at r with starting positions from 0 to l-1 will be valid because they all contain the substring s[l...r].

Why does this work? Consider a window [l, r] where some character c appears exactly k times. Any substring that starts before l and ends at r will contain all occurrences of c in the window [l, r], so it will also have at least k occurrences of c.

The sliding window maintains the invariant: the window [l, r] is the minimal window ending at r that contains a character appearing at least k times. As we extend r, we shrink from the left (increment l) whenever we have a character with count ≥ k, ensuring we always have the minimal such window. The number of valid substrings ending at each position r is exactly l, representing all possible starting positions from 0 to l-1.

Learn more about Sliding Window patterns.

Solution Approach

We implement a sliding window solution using two pointers to efficiently count valid substrings.

Data Structures:

• cnt: A Counter (hash map) to track the frequency of each character in the current window [l, r]
• l: Left pointer of the sliding window, initially set to 0
• ans: Accumulator for the total count of valid substrings

Algorithm Steps:

1. Enumerate the right endpoint: We iterate through each character c in the string s, treating each position as a potential right endpoint r of substrings.

2. Expand the window: For each character at position r, we add it to our frequency counter: cnt[c] += 1

3. Shrink from the left when condition is met: When the current character c appears at least k times in the window (cnt[c] >= k), we need to find the minimal window ending at r that satisfies our condition. We do this by:

   • Removing characters from the left: cnt[s[l]] -= 1
   • Moving the left pointer rightward: l += 1
   • Continuing this process while cnt[c] >= k

4. Count valid substrings: After adjusting the window, l represents the leftmost position where the window [l, r] no longer contains any character appearing ≥ k times. This means all substrings ending at r with starting positions from 0 to l-1 are valid. We add l to our answer.

Example Walkthrough:

Consider s = "aaab" and k = 2:

• When r = 0 (first 'a'): cnt['a'] = 1, no character has count ≥ 2, so l = 0, add 0
• When r = 1 (second 'a'): cnt['a'] = 2, now 'a' appears 2 times
  • Shrink: remove s[0], cnt['a'] = 1, move l = 1
  • Now 'a' appears only once in window [1, 1]
  • Add l = 1 to answer (substring "aa" starting at index 0 is valid)
• When r = 2 (third 'a'): cnt['a'] = 2 again
  • Shrink: remove s[1], move l = 2
  • Add l = 2 to answer (substrings "aaa" and "aa" starting at indices 0 and 1 are valid)
• When r = 3 ('b'): cnt['b'] = 1, no shrinking needed
  • Add l = 2 to answer (substrings ending at 'b' starting from indices 0 and 1 are valid)

The time complexity is O(n) where n is the length of the string, as each character is added and removed from the window at most once. The space complexity is O(min(n, 26)) for storing character frequencies.

Example Walkthrough

Let's trace through the algorithm with s = "abba" and k = 2.

Initial State:

• l = 0, ans = 0, cnt = {}

Step 1: r = 0, c = 'a'

• Add 'a' to counter: cnt = {'a': 1}
• cnt['a'] = 1 < 2, so no shrinking needed
• Add l = 0 to answer → ans = 0
• Window [0,0] = "a" has no character appearing ≥ 2 times

Step 2: r = 1, c = 'b'

• Add 'b' to counter: cnt = {'a': 1, 'b': 1}
• cnt['b'] = 1 < 2, so no shrinking needed
• Add l = 0 to answer → ans = 0
• Window [0,1] = "ab" has no character appearing ≥ 2 times

Step 3: r = 2, c = 'b'

• Add 'b' to counter: cnt = {'a': 1, 'b': 2}
• cnt['b'] = 2 ≥ 2, so we need to shrink!
  • Remove s[0] = 'a': cnt = {'a': 0, 'b': 2}
  • Move l = 1
  • Check: cnt['b'] = 2 ≥ 2, continue shrinking
  • Remove s[1] = 'b': cnt = {'a': 0, 'b': 1}
  • Move l = 2
  • Check: cnt['b'] = 1 < 2, stop shrinking
• Add l = 2 to answer → ans = 2
• This counts substrings "abb" (starting at 0) and "bb" (starting at 1)

Step 4: r = 3, c = 'a'

• Add 'a' to counter: cnt = {'a': 1, 'b': 1}
• cnt['a'] = 1 < 2, so no shrinking needed
• Add l = 2 to answer → ans = 4
• This counts substrings "abba" (starting at 0) and "bba" (starting at 1)

Final Result: ans = 4

The valid substrings are:

1. "abb" - 'b' appears 2 times
2. "abba" - 'b' appears 2 times
3. "bb" - 'b' appears 2 times
4. "bba" - 'b' appears 2 times

The algorithm correctly identifies that once we have a minimal window with a character appearing ≥ k times, all extensions to the left create valid substrings.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string s. This is because we iterate through each character in the string exactly once with the outer loop. While there is an inner while loop, each character can only be processed by the left pointer l at most once throughout the entire execution (the left pointer only moves forward and never backwards), making the amortized cost O(1) per character. Therefore, the total operations are bounded by 2n, which simplifies to O(n).

The space complexity is O(|Σ|), where Σ is the character set. Since the problem deals with lowercase letters (as implied by the reference answer), we have |Σ| = 26. The Counter object cnt stores at most one entry for each distinct character that appears in the string, which is bounded by the size of the alphabet. Therefore, the space complexity is O(26) = O(1) in terms of constant space, or more precisely O(|Σ|) when considering the character set size as a parameter.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Misunderstanding the Counting Logic

The Problem: A common mistake is thinking that when we find a valid window [l, r] where some character appears at least k times, we should count ALL substrings within that window. This leads to incorrect counting logic like:

# INCORRECT approach
if char_count[char] >= k:
    # Wrong: counting all substrings in current window
    total_substrings += (right - left + 1)

Why It's Wrong: This counts substrings that may not contain any character appearing k times. For example, if window [0, 3] has character 'a' appearing 3 times, the substring [1, 1] within it might only have 'b', which doesn't satisfy our condition.

The Correct Understanding: The solution counts substrings that END at position right. When we shrink the window to [left, right] where no character appears ≥ k times, it means all substrings ending at right and starting from positions 0 to left-1 ARE valid (because they contain the previous valid window as a subset).

Pitfall 2: Incorrect Window Shrinking Condition

The Problem: Some might try to shrink the window while ANY character in the window has count ≥ k:

# INCORRECT shrinking condition
while any(count >= k for count in char_count.values()):
    char_count[s[left]] -= 1
    left += 1

Why It's Wrong: This is inefficient and conceptually incorrect. We only need to shrink when the NEWLY ADDED character at position right reaches count k. This ensures we find the minimal valid window ending at each position.

The Correct Approach: Only shrink while the current character being processed has count ≥ k:

while char_count[char] >= k:  # 'char' is s[right]
    char_count[s[left]] -= 1
    left += 1

Pitfall 3: Off-by-One Error in Counting

The Problem: After shrinking the window, some might incorrectly count:

# INCORRECT counting
total_substrings += left + 1  # Wrong!

Why It's Wrong: After shrinking, left points to the first position where the window [left, right] does NOT satisfy our condition. This means valid substrings ending at right start from indices 0 to left-1, which gives us exactly left valid substrings, not left + 1.

The Correct Count:

total_substrings += left  # Correct: counts substrings from [0,right] to [left-1,right]